The moment of inertia of a uniform semi-circular disc about an axis passing through its centre of mass and perpendicular to its plane is (Mass of this disc is M and radius is R ).

Question

The moment of inertia of a uniform semi-circular disc about an axis passing through its centre of mass and perpendicular to its plane is (Mass of this disc is M and radius is R ). (A) (( textMR)2/2) - textM ((2 textR/π ))2 (B) (( textMR)2/2) - textM ((4 textR/3 π ))2 (C) (( textMR)2/2) + textM ((4 textR/3 π ))2 (D) (( textMR)2/2) + textM ((2 textR/π ))2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/947f87f5ed658da02579270e94ee137e-.png" /> ( textI)0 = ((( textMR)2/2)) textI0 = textI textcm + textMd2 (( textMR)2/2) = ( textI) textcm + textM ((4 textR/3 π ))2 â ( textI) textcm = (( textMR)2/2) - textM ((4 textR/3 π ))2 ( textI) textcm = [(( textMR)2/2) - textM ((4 textR/3 π ))2]

Q. The moment of inertia of a uniform semi-circular disc about an axis passing through its centre of mass and perpendicular to its plane is (Mass of this disc is $M$ and radius is $R$ ).

$\frac{( MR ) ^{2}}{2} - M (\frac{2 R}{π})^{2}$

$\frac{( MR ) ^{2}}{2} - M (\frac{4 R}{3 π})^{2}$

$\frac{( MR ) ^{2}}{2} + M (\frac{4 R}{3 π})^{2}$

$\frac{( MR ) ^{2}}{2} + M (\frac{2 R}{π})^{2}$

$(I)_{0} = (\frac{( MR ) ^{2}}{2})$
$I_{0} = I_{cm} + Md^{2}$
$\frac{( MR ) ^{2}}{2} = (I)_{cm} + M (\frac{4 R}{3 π})^{2}$
$\Rightarrow (I)_{cm} = \frac{( MR ) ^{2}}{2} - M (\frac{4 R}{3 π})^{2}$
$(I)_{cm} = [\frac{( MR ) ^{2}}{2} - M (\frac{4 R}{3 π})^{2}]$

Q. The moment of inertia of a uniform semi-circular disc about an axis passing through its centre of mass and perpendicular to its plane is (Mass of this disc is M and radius is R ).

2(MR)2​−M(π2R​)2

2(MR)2​−M(3π4R​)2

2(MR)2​+M(3π4R​)2

2(MR)2​+M(π2R​)2